Problem: An astronomer knows the distances from herself to stars $A$ and $B$, as well as the distance between them. The distances are $450$, $400$, and $90$ light years ( $\text{l.y.}$ ) respectively. If the astronomer's telescope is currently pointed at star $A$, how many degrees must she rotate her telescope to see star $B$ ? Do not round during your calculations. Round your final answer to the nearest degree.
Answer: Converting the problem into geometrical terms Our problem can be modeled by the following triangle $\triangle ABC$, where we want to find $\angle C=\theta$. $C$ $\theta$ $\;\;\;\;\;\;\;\;\;\;\;\;A$ $B$ $450\text{ l.y.}$ $400\text{ l.y.}$ $90\text{ l.y.}$ Since we are given three side lengths, we can use the law of cosines. Using the law of cosines The law of cosines gives the following equation. $(AB)^2=(AC)^2+(BC)^2-2AC\!\cdot\! BC\!\cdot\!\cos(C)$ Solving the above equation for $\cos(C)$ gives the following equation. $\begin{aligned} \cos(C)&=\dfrac{(AC)^2+(BC)^2-(AB)^2}{2AC\!\cdot\! BC}\\\\ \cos(\theta)&=\dfrac{450^2+400^2-90^2}{2\cdot 450\cdot 400} \gray{\text{Substitute}}\\\\ \cos(\theta)&=\dfrac{354{,}400}{360{,}000}\\\\ \theta&=\cos^{-1}\left(\dfrac{354{,}400}{360{,}000}\right)\\\\ \theta&\approx 10^\circ \end{aligned}$ The answer The astronomer must turn her telescope by $10^\circ$.